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IiJ 0 ~ nk s 1/k and n~ = 0. E-Qi! > 1/k for a'Ll E in k PROOF. ~ 1 be the canonical ONB of ~ and let Pk k ] ]- k k-j (1/k)}:. t9e. t9e. (n = 1,2, ••• ). ld = rank F ?? > ~ 1/k for (at least) one value of j, 1 ::; j ::; k. , = 0 for 1 s j s i s k. >I?? , &k_ ~ 1/k. , F admits an upper triangular matrix with 1/k in all the diagonal entries with respect to the ONB {fj}j!

R=1 '' · 1/k+1/n, 25 we see that 15k_ s; 1/k. On the other hand, if F o (F) c { O,l} and € k trace F = Lj=l = rank F ?? > ~ 1/k for (at least) one value of j, 1 ::; j ::; k. , = 0 for 1 s j s i s k. >I?? , &k_ ~ 1/k. , F admits an upper triangular matrix with 1/k in all the diagonal entries with respect to the ONB {fj}j! 1 • But the second possibility implies that o(F) = {1/k}, a contradiction (recall that k ~ 2). Hence, IIF- Qll > ~Sk_ = 1/k for all F in E(~k) and all Q in N(~k).

R=1 '' · 1/k+1/n, 25 we see that 15k_ s; 1/k. On the other hand, if F o (F) c { O,l} and € k trace F = Lj=l = rank F ?? > ~ 1/k for (at least) one value of j, 1 ::; j ::; k. , = 0 for 1 s j s i s k. >I?? , &k_ ~ 1/k. , F admits an upper triangular matrix with 1/k in all the diagonal entries with respect to the ONB {fj}j! 1 • But the second possibility implies that o(F) = {1/k}, a contradiction (recall that k ~ 2). Hence, IIF- Qll > ~Sk_ = 1/k for all F in E(~k) and all Q in N(~k).

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